Every Weakly Compact Set in X is Norm Bounded

Theorem 1: Let $X$ be a normed linear space and let $K \subset X$. If $K$ is weakly compact then $K$ is norm bounded.

• Proof: Suppose that $K$ is weakly compact.

• Since $X$ is a normed space, for each $k \in K$ by one of the corollaries to Hahn-Banach theorem on the Corollaries to the Hahn-Banach Theorem there exists a $f_k \in X^*$ such that $\| f_k \| = 1$ and $f_k(k) = \| k \|$.

• Since each $f_k \in X^*$ and since $K$ is compact, we have by the Extreme Value theorem that $f_k$ attains a maximum and minimum on $K$, that is, $f_k(K)$ is bounded. So for each $k \in K$:

(1)

\begin{align} \quad \sup_{t \in K} | f_k(t) | = \sup_{t \in K} | \hat{t}(f_k) \| < \infty \end{align}

• Since $X^* = B(X, \mathbb{R})$ is always a Banach space, by The Uniform Boundedness Principle we have that:

(2)

\begin{align} \quad \sup_{t \in K} \| \hat{t} \| < \infty \end{align}

• But $\| \hat{t} \| = \| t \|$, and so:

(3)

\begin{align} \quad \sup_{t \in K} \| t \| < \infty \end{align}

• Thus $K$ is norm bounded. $\blacksquare$